How can we deal with a network branch which is part of two networks each with a source? R3 is carrying current supplied by each battery

Transcription

1 Network nalysis ims: Consolidate use of KCL in circuit analysis. Use Principle of Superposition. Learn basics of Node Voltage nalysis (uses KCL) Learn basics of Mesh Current nalysis (uses KVL) Lecture 6 1 Superposition How can we deal with a network branch which is part of two networks each with a source? R1 R2 V1 R V2 R is carrying current supplied by each battery Principle of superposition: The current in a branch of a linear circuit supplied from several sources is equal to the sum of the currents in that component provided by each source alone with the other sources reduced to zero Reduced to zero means: Voltage sources short circuit (V=0, R=0) Current sources open circuit (I=0, R= ) Lecture 6 2 1

2 Example#1 2 2 Ω 14 Ω 8 V I 2 II? 1 I? Lecture 6 Example #1 2 2 Ω 14 Ω 8 V I? Stage 1: Current source to zero (open circuit): I 1 = 8V/16 Ω = Ω 14 Ω 8 V I1 Stage 2: Voltage source to zero (short circuit): R B = 14 2 = 28/16 Ω V B = R B x 2 = 56/16 V I 2 = V B / 14 = I2 B 2 Ω 14 Ω Stage : Superposition: I = I 1 + I 2 = 0.75 Lecture 6 5 2

3 pplications Superposition tells us that branch currents and voltages are always proportional to the driving currents or voltages. This means that resistor networks are LINER CIRCUITS Network analysis We know that we can replace complex networks of sources and resistors by single sources (Thévenin, Norton), so we can use superposition to calculate the effects of networks (or sub-networks) feeding the same load Sensitivity analysis How much does a branch current or node voltage change when one of the sources changes? (This becomes very important when you have C signals in your network) Working back Do a calculation with an easy number and then scale the result to get the actual answer you want for a difficult number Lecture 6 6 Working back (or the unit method) simple example: Calculate the voltage required to give 17 m in the 1Ω resistor 5 V? 17 m 1 The working back approach: calculate the voltage required to give 1 in the resistor: V = 1 x (2 + 1) = V Current in 5Ω = 1 + V / = 2 V = V + 5 x 2 = 1V So 1V gives 1 To get 17m, the superposition principle then tells us that we need 1V x (17 m / 1 ) = 221 mv. 2 Lecture 6 7

4 Superposition Example#2 Find the current flowing through B I 6 42V V B 5 4Ω Ω Lecture 6 8 Superposition Example#2 Useful when there are multiple sources: Find the current flowing through B = = I R n eq 2 = = I R 6 n 42 V 6 B 5 1: Reduce voltage source to zero: I 1 6 B I1 = -5/ 0 V 5 Lecture

5 Network analysis using superposition 2: Reduce current source to zero: I 2 6 B 42 V I2 = 14/ 0 So the total current (from superposition) is I = I1 + I2 = -5/ + 14/ = 9/ = Lecture 6 11 General network analysis So far, most of the methods we have considered are tricks to simplify the analysis of a network (e.g. by bundling up resistors and sources, turning sources off one by one) Sometimes we wish to know the current and voltage at every branch/node in any general network. For this we fall back on KCL and KVL. There are two different approaches: Node voltage analysis: Get an equation for each node voltage in the network and use these to solve for the branch currents Mesh current analysis: Get an equation for each branch current in the network and use these to solve for node voltages Lecture 6 5

6 Node Voltage nalysis Most appropriate when There are only current sources in the network OR There are fewer nodes than meshes Method: systematic use of KCL at all but one of the nodes in a network. (The node left out is the reference or ground node) The currents for KCL are calculated in terms of the node voltages and branch resistances If there are N nodes, you get N-1 equations for node voltage, which allows you to obtain all the unknown voltages s always, it is essential to get the signs and polarities right Lecture 6 1 n example: 4 nodes, 4 meshes, current sources only: Use node voltage analysis 1 v v0 = 0 v2 2 Label nodes 0, 1 and 2. Node 0 is the reference (ground), so we will try to find v1 and v2 Choose a sign convention for KCL (e.g. currents leaving are positive) and mark this on the diagram Branch currents are then given either by the current source value or Ohm s law on the resistors (voltage difference / resistance) Lecture

7 n example: v1 4 v2 KCL at node 1: v1 v0 v1 v = KCL at node 2: v2 v0 v2 v = v0 = v1 v2 = v1 + v2 = Lecture 6 15 Matrix formulation 5 1 v1 v2 = v1 + v2 = These simultaneous equations can be solved to give: v1 = 64/9 V and v2 = -40/9 V nd the unknown currents can then be found The equations can also be written in MTRIX form: I is the matrix of the sum of source currents at each node V is the matrix of node voltages G is the conductance matrix Diagonal terms are the sum of conductances at each node Non-diagonal terms are the conductance joining each pair of nodes v1 5 = 1 1 v or GV = I Matrix algebra is a very powerful tool for handling large numbers of simultaneous equations, and this gives us a route to scale up the analysis to a general number of nodes and to write computer programs to do this! Lecture

8 Mesh current analysis Most appropriate when There are only voltage sources in the network OR There are fewer meshes or loops than nodes Method: systematic use of KVL to obtain the current in each mesh in a network. Identify each mesh (1 to N) and mark clockwise mesh currents (i n ) Identify each branch (1 to M) Branch currents of shared branches are the difference of the mesh currents 5Ω 1Ω 4Ω 6 V I1 6Ω I2 Ω I 2Ω Lecture 6 17 Example 5 nodes, meshes, only voltage sources: Use mesh current analysis. 6 V 5Ω 1Ω 4Ω I1 6Ω I2 Ω I 2Ω KVL in mesh 1: 6+ 5i + 6( i i ) = 0 mesh 2 : ( i i ) + i + ( i i ) = mesh : ( i i ) + 4i + 2i = i1 6i2 + 0i 6i1+ 10i2 i = 6 = 0 0 i i + 9i = This can be solved by substitution to give i 1 =9, i 2 = 6 and i = 2 The branch currents and node voltages then follow Lecture

9 nother matrix: 11 i 6i + 0i = i + 10i i = i i + 9i = i i = i 0 RI = V V is the matrix of the sum of voltage sources in each mesh I is the matrix of mesh currents R is the resistance matrix Diagonal elements are sum of resistance in each mesh Off-diagonal elements are coupling resistance between meshes (e.g. meshes 1 and 2 are coupled by 6Ω resistor, meshes 1 and have no coupling) Once again, matrix techniques form the basis of a general approach to solving any arbitrary network Lecture 6 19 Network analysis software These programs allow you to define a network of nodes with components linking them. Not just resistors and sources, but capacitors, inductors, dependent sources or any other components that can be defined as an equivalent circuit (e.g. FETs, transistors). The programs build the conductance, resistance and source matrices for the network and use these to determine mesh currents and node voltages. Many features such as time and frequency dependence, graphs of everything against everything else, etc Lecture

10 Network analysis software SPICE : The industry standard professional package used for modelling everything from complex integrated circuits to telephone networks. The full version is massively expensive, but it has spawned a whole family of SPICE-like packages, some of which have limited student or demo versions. Microcap PSPICE: This may be available on the Physics workstations. Otherwise try: Lecture 6 21 Summary of techniques for network analysis Technique Ohm s law Series and parallel Source transformation (Norton, Thévenin) Superposition Node voltage analysis Mesh current analysis Loop current analysis Typical application Current and voltage in simple loops Simplifying networks with ONLY resistors Calculating the current/voltage at only one branch/node of a complex network when you are not interested in what is happening in the rest Calculating current/voltage in a branch fed with multiple sources Calculate the current and voltage at every node and branch in a network Calculate the current and voltage at every node and branch in a network Lecture